Answer:
The final volumetric flow rate will be "76.4 m³/s".
Explanation:
The given values are:
[tex]\dot{m_{1}}=10 \ kg/s[/tex]
[tex]\dot{m_{2}}=20 \ Kg/s[/tex]
[tex]T_{1}=293 \ K[/tex]
[tex]T_{2}=313 \ K[/tex]
[tex]P_{1}=P_{2}=P_{3}=10 \ bar[/tex]
As we know,
⇒ [tex]E_{in}=E_{out}[/tex]
[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]
[tex]e_{1}\dot{v_{1}}h_{1}+e_{2}\dot{v_{2}}h_{2}=e_{3}\dot{v_{3}}h_{3}[/tex]
[tex]\frac{P_{1}}{RP_{1}}\dot{v_{1}} \ C_{p}T_{1}+ \frac{P_{2}}{RP_{2}}\dot{v_{2}} \ C_{p}T_{1}=\frac{P_{3}}{RP_{3}}\dot{v_{3}} \ C_{p}T_{3}[/tex]
⇒ [tex]\dot{v_{3}}=\dot{v_{1}}+\dot{v_{2}}[/tex]
[tex]=\frac{\dot{m_{1}}}{e_{1}}+\frac{\dot{m_{2}}}{e_{2}}[/tex]
On substituting the values, we get
[tex]=\frac{10}{10\times 10^5}\times 8314\times 293+\frac{20\times 8314\times 313}{10\times 10^5}[/tex]
[tex]=76.4 \ m^3/s[/tex]
A long corridor has a single light bulb and two doors with light switch at each door. design logic circuit for the light; assume that the light is off when both switches are in the same position.
Answer and Explanation:
Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:
A B C (output)
0 0 0
0 1 1
1 0 1
1 1 0
The logic circuit is shown below
C = A'B + AB'
If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacuum. The two sides are separated by a piston that is initially held in place by the pins. The pins are removed and the gas suddenly expands until it hits the stops. What happens to the internal energy of the gas?
a. internal energy goes up
b. internal energy goes down
c. internal energy stays the same
d. we need to know the volumes to make the calculation
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.
A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lane is provided on the local road to allow vehicles from the ramp to turn right onto the local road without stopping. The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle. Determine the width of the turning roadway if the design vehicle is a single-unit truck. Use 0.08 for superelevation.
Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
[tex]R = \dfrac{u^2}{15(e+f_s)}[/tex]
where;
R= radius
[tex]f_s[/tex] = coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
[tex]f_s[/tex] = 0.20
So;
[tex]R = \dfrac{30^2}{15(0.08+0.20)}[/tex]
[tex]R = \dfrac{900}{15(0.28)}[/tex]
[tex]R = \dfrac{900}{4.2}[/tex]
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
The value of an SMT capacitor is signified by a
Answer:
Working volttage
Explanation:
SMT electrolytic capacitors are marked with working voltage. The value of these capacitors is measured in micro farads. It is a surface mount capacitor which is used for high volume manufacturers. They are small lead less and are widely used. They are placed on modern circuit boards.
For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in. An external load of 80 kips is applied to the entire joint. Assume the load is equally distributed to all the bolts. It has been determined to use 1/2 in- 13 UNC grade 8 bolts with rolled threads. Assume the bolts are preloaded to 75% of the proof load. Clearly state any assumptions.
(a) Determine the yielding factor of safety,
(b) Determine the overload factor of safety,
(c) Determine the factor of safety baserd on joint seperation.
Answer:
nP ≈ 4.9 nL = 1.50Explanation:
GIVEN DATA
external load applied (p) = 85 kips
bolt stiffness ( Kb ) = 3(10^6) Ibf / in
Member stiffness (Km) = 12(10^6) Ibf / in
Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8
Number of bolts = 6
assumptions
for unified screw threads UNC and UNF
tensile stress area ( A ) = 0.1419 in^2
SAE specifications for steel bolts for grade 8
we have
Minimum proff strength ( Sp) = 120 kpsi
Minimum tensile strength (St) = 150 Kpsi
Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips
Given the following values
Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip
Preload stress
αi = 0.75Sp = 0.75 * 120 = 90 kpsi
stiffness constant
C = [tex]\frac{Kb}{Kb + Km}[/tex] = [tex]\frac{3}{3+2}[/tex] = 0.2
A) yielding factor of safety
nP = [tex]\frac{sPAt}{Cp + Fi}[/tex] = [tex]\frac{120* 0.1419}{0.2*14.17 + 12.771}[/tex]
nP = 77.028 / 15.605 = 4.94 ≈ 4.9
B) Determine the overload factor safety
[tex]nL = \frac{SpAt - Fi}{CP}[/tex] = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17
= 17.028 - 12.771 / 2.834
= 1.50